Beginning and Intermediate Algebra - Wallace Math Courses ...

4(30 −n)+2.5n=105 Substitute into untouched equation
120 − 4n + 2.5n=105 Distribute
120 − 1.5n=105 Combine like terms
− 120 − 120 Subtract 120 from both sides
− 1.5n=−15 Divide both sides by − 1.5
− 1.5 − 1.5
n=10 We have ourn, 10 lbs of nuts
c = 30 − (10) Plug into c=equation to find c
c=20 We have ourc, 20 lbs of chocolate
10 lbs of nuts and 20 lbs of chocolate Our Solution
With mixture problems we often are mixing with a pure solution or using water
which contains none of our chemical we are interested in. For pure solutions, the
percentage is 100% (or 1 in the table). For water, the percentage is 0%. This is
shown in the following example.
Example 195.
A solution of pure antifreeze is mixed with water to make a 65% antifreeze solution.
How much of each should be used to make 70 L?
Amount Part Final
Antifreeze a 1
Water w 0
Final 70 0.65
Amount Part Final
Antifreeze a 1 a
Water w 0 0
Final 70 0.65 45.5
We use a andw for our variables. Antifreeze
is pure, 100% or 1 in our table, written asa
decimal. Water has no antifreeze, its
percentage is 0. We also fill in the final percent
Multiply to find final amounts
a +w = 70 First equation comes from first column
a = 45.5 Second equation comes from second column
(45.5) +w = 70 We havea, plug into to other equation
− 45.5 − 45.5 Subtract 45.5 from both sides
w = 24.5 We have ourw
45.5L of antifreeze and 24.5L of water Our Solution
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